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likoan [24]
2 years ago
8

A brick of aluminum weighs 19,745 grams and has the following dimensions: length 29 cm, height 15 cm, and width 51 mm. What is t

he approximate density of the aluminum? Round to the nearest tenth.
Mathematics
1 answer:
garri49 [273]2 years ago
3 0

If the volume of the aluminum brick will be 2218.5 cubic cm. Then the approximate density of the aluminum will be 8.9 grams per cubic cm.

<h3>What is density?</h3>

Density is defined as the mass per unit volume. It is an important parameter in order to understand the fluid and its properties. Its unit is kg/m³.

The mass and density relation is given as

Density = mass/volume

A brick of aluminum weighs 19,745 grams and has the following dimensions: length 29 cm, height 15 cm, and width 51 mm.

Then the approximate density of the aluminum will be

The volume of the aluminum brick will be

V = 29 x 15 x 5.1

V = 2218.5 cubic cm

Then the density will be

D = 19,745 / 2218.5

D = 8.900 gram per cubic cm

To learn more about the density refer to the link;

brainly.com/question/952755

#SPJ1

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Answer:

The probability of Thomas inviting Madeline to the party over the phone is 0.143.

Step-by-step explanation:

Consider the tree diagram below.

The events are denoted as follows:

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<em>B</em> = Thomas call Madeline on the phone

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The conditional probability of event <em>U</em> given that another events <em>V</em> has already occurred is:

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The law of total probability states that:

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In this case we need to determine the probability that Thomas invites Madeline to the party over the phone, i.e. P (B|X).

Use the law of total probability to determine the value of P (X) as follows:

P(X) = P(X|A)P(A)+P(X|B)P(B)

         =(0.90\times 0.80)+(0.60\times 0.20)\\=0.72+0.12\\=0.84

Compute the value of P (B|X) as follows:

P(B|X)=\frac{P(X|B)P(B)}{P(X)}

             =\farc{0.60\times 0.20}{0.84}\\\\=0.14286\\\\\approx 0.143

Thus, the probability of Thomas inviting Madeline to the party over the phone is 0.143.

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